A glass of radius `10^(-3)` and density `2000 kg m^(-3)` fall in a jar filled with oil of density `800 kg m^(-3)` .The terminal velocity is found to be 1 cm/s. Calculate the coefficient of viscosity of oil

Explanation

To calculate the coefficient of viscosity of oil when a glass sphere falls through it, we can use the formula for terminal velocity in a viscous fluid. The formula is given by:

\[ v_t = \frac{2}{9} \cdot \frac{r^2 \cdot g}{\eta} \cdot (\rho_s - \rho_l) \]

Where:
- \( v_t \) = terminal velocity
- \( r \) = radius of the sphere
- \( g \) = acceleration due to gravity
- \( \eta \) = coefficient of viscosity
- \( \rho_s \) = density of the sphere (glass)
- \( \rho_l \) = density of the liquid (oil)

### Step 1: Rearranging the formula
We need to solve for the coefficient of viscosity \( \eta \). Rearranging the formula gives us:

\[ \eta = \frac{2}{9} \cdot \frac{r^2 \cdot g}{v_t} \cdot (\rho_s - \rho_l) \]

### Step 2: Substituting the known values
Now we will substitute the known values into the equation:
- Radius \( r = 10^{-3} \) m
- Density of glass \( \rho_s = 2000 \) kg/m³
- Density of oil \( \rho_l = 800 \) kg/m³
- Terminal velocity \( v_t = 1 \) cm/s = \( 0.01 \) m/s (conversion from cm/s to m/s)
- Acceleration due to gravity \( g = 9.8 \) m/s²

Substituting these values into the equation:

\[ \eta = \frac{2}{9} \cdot \frac{(10^{-3})^2 \cdot 9.8}{0.01} \cdot (2000 - 800) \]

### Step 3: Calculating the differences in density
Calculate \( \rho_s - \rho_l \):

\[ \rho_s - \rho_l = 2000 - 800 = 1200 \text{ kg/m}^3 \]

### Step 4: Plugging in the values
Now substituting this back into the equation:

\[ \eta = \frac{2}{9} \cdot \frac{(10^{-6}) \cdot 9.8}{0.01} \cdot 1200 \]

### Step 5: Simplifying the expression
Calculating the terms step by step:

1. \( (10^{-3})^2 = 10^{-6} \)
2. \( \frac{9.8}{0.01} = 980 \)
3. Now substituting these values:

\[ \eta = \frac{2}{9} \cdot 10^{-6} \cdot 980 \cdot 1200 \]

### Step 6: Final calculation
Calculating the coefficient of viscosity:

\[ \eta = \frac{2 \cdot 980 \cdot 1200 \cdot 10^{-6}}{9} \]

Calculating \( 2 \cdot 980 \cdot 1200 = 2352000 \):

\[ \eta = \frac{2352000 \cdot 10^{-6}}{9} = \frac{2.352 \cdot 10^3}{9} \approx 261.33 \cdot 10^{-3} \text{ Pa.s} \approx 0.261 \text{ Pa.s} \]

### Conclusion
Thus, the coefficient of viscosity of the oil is approximately:

\[ \eta \approx 0.261 \text{ Pa.s} \]